most recent 30 from http://mathoverflow.net 2013-05-26T07:38:49Z http://mathoverflow.net/feeds/question/5196 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/5196/on-algebraic-field-extensions Nick Gill 2009-11-12T14:51:19Z 2009-11-17T01:18:08Z

<p>Let $L:K$ be a field extension. Let $A$ be a set of elements in $L$, all of which are algebraic over $K$. Construct the field extension $M=K(A)$. I have two questions:</p> <p>[1] Is $M:K$ an algebraic field extension?</p> <p>[2] Take $\beta\in M$ where $\beta$ is algebraic over $K$. Then $K(\beta):K$ is a finite extension. Can I assert that $\beta$ lies in a finite extension $K(a1,..., an)$ where $a1,..., an\in A$?</p> <p>Both questions are trivial when $A$ is finite. So assume that $A$ is infinite; indeed assume that $[M:K]$ is infinite. Now what?</p>

http://mathoverflow.net/questions/5196/on-algebraic-field-extensions/5763#5763 Greg Kuperberg 2009-11-17T01:18:08Z 2009-11-17T01:18:08Z

<p>The question was a bit confusing because it was not clear what was meant as the definition of "the field extension $K(A)$". Nick says that "the smallest field containing" is the definition that he wanted, to make the question non-trivial. (But still routine, in my opinion.)</p> <p>Let's agree then that both (1) and (2) are answered by this summary: The smallest field that contains $K$ and $A$ is the set of rational expressions in $A$ with coefficients in $K$. An element $\beta$ of this field $M$ can only use finitely many elements of $A$, so $\beta$ lies in the subfield generated by them. Moreover, $\beta$ is algebraic by the lemma that the sum or product of two algebraic numbers is algebraic, and the reciprocal of an algebraic number is algebraic.</p> <p>I'm putting this summary here to hopefully take the question off of the unanswered stack.</p>